Is there a group with countably many subgroups, but is not countable in ZF?

Solution 1:

I believe the answer is yes, as follows:

Start with a model of ZF+atoms, $M$, with a set of atoms $A$ which forms a group isomorphic to $\mathbb{D}/\mathbb{Z}$, where $\mathbb{D}$ is the set of dyadic fractions: $\mathbb{D}=\{{p\over 2^k}: p, k\in\mathbb{Z}\}$. Let $G$ be the group of automorphisms of $A$, and consider the symmetric submodel $N$ of $M$ corresponding to the filter of finite supports on $G$. Then in $N$, $A$ is no longer countable, since there are nontrivial automorphisms of $\mathbb{D}/\mathbb{Z}$ fixing arbitrary finite sets; but the only subgroups of $\mathbb{D}/\mathbb{Z}$, other than the whole thing, are those of the form $$\{x: 2^kx=0\}$$ for some fixed $k\in\mathbb{N}$. This provides an explicit bijection - in the original universe, $M$ - between the subgroups of $A$ and $\omega$. Now, passing to $N$, we get no additional subgroups of $A$, and the map described above is symmetric; so in $N$, $A$ has only countably many subgroups.

Meanwhile, the statements "$A$ is uncountable" and "$A$ has countably many finitely generated subgroups" are each "bounded," so we may apply the Jech-Sochor theorem to push this construction into the ZF-setting.